compare sums of individuals columns and return name of max and min column pandas-CodePudding

I have a df something like this

     date         mon       tue       wed       thu         fri       sat        sun
  01-01-2022       2         3         5         7           8         1          0
  02-01-2022       3         4         7         6           3         0          4
  03-01-2022       4         8         7         9           1         2          5
  04-01-2022       5         2         1         1           8         1          2
  05-01-2022       6         1         9         3           7         1          1

my task is to find the sum of each column and compare them to each other and return the name of the column that has the highest and lowest sum. So in this case, when I compare sums of each column, I have wed column to be at max sum (29) and sat to be at min sum (5). So my expected output is printing this information:

   max number is seen on wed and min number is seen on sat.

can someone please help me with an efficient way of doing this? Much appreciated

CodePudding user response：

You can use set_index, sum and agg:

df.set_index('date').sum().agg(['idxmin', 'idxmax'])

output:

idxmin    sat
idxmax    wed
dtype: object

As a string:

s = df.set_index('date').sum().agg(['idxmin', 'idxmax'])

print(f"max number is seen on {s['idxmax']} and min number is seen on {s['idxmax']}.")

output:

max number is seen on wed and min number is seen on wed.